The separation energy of the electron present in the shell n = 3 is 1.51 eV. What is the energy in the first excited state -

Options

A. `-1.51eV`
B. `-3.4 eV`
C. `+1.51eV`
D. `+3.4eV`

Explanation

To solve the problem, we need to determine the energy of the electron in the first excited state (n = 2) given the separation energy of the electron in the shell n = 3 (which is 1.51 eV).

### Step-by-Step Solution:

1. **Understand the given information**:
- The separation energy of the electron at n = 3 is given as 1.51 eV.
- The formula for the energy of an electron in a hydrogen-like atom is given by:
\[ E_n = -\frac{13.6 \, Z^2}{n^2} \text{ eV} \]
- Here, Z is the atomic number, and n is the principal quantum number.

2. **Relate the separation energy to the energy levels**:
- The separation energy (ΔE) can be expressed as:
\[ \Delta E = E_{\infty} - E_n \]
- For n = 3, we have:
\[ E_{\infty} = 0 \text{ eV (energy at infinite distance)} \]
- Thus, we can rewrite the equation as:
\[ \Delta E = 0 - E_3 = -E_3 \]
- Therefore:
\[ E_3 = -1.51 \text{ eV} \]

3. **Calculate Z using the energy at n = 3**:
- Substitute n = 3 into the energy formula:
\[ -1.51 = -\frac{13.6 \, Z^2}{3^2} \]
- Simplifying gives:
\[ -1.51 = -\frac{13.6 \, Z^2}{9} \]
- Rearranging gives:
\[ 1.51 = \frac{13.6 \, Z^2}{9} \]
- Multiplying both sides by 9:
\[ 1.51 \times 9 = 13.6 \, Z^2 \]
- This simplifies to:
\[ 13.59 = 13.6 \, Z^2 \]
- Dividing both sides by 13.6:
\[ Z^2 = \frac{13.59}{13.6} \approx 1 \]
- Taking the square root gives:
\[ Z \approx 1 \]
- This indicates we are dealing with a hydrogen atom.

4. **Calculate the energy in the first excited state (n = 2)**:
- Now we can find the energy at n = 2:
\[ E_2 = -\frac{13.6 \, Z^2}{2^2} \]
- Substituting Z = 1:
\[ E_2 = -\frac{13.6 \cdot 1^2}{4} = -\frac{13.6}{4} = -3.4 \text{ eV} \]

### Final Answer:
The energy in the first excited state (n = 2) is **-3.4 eV**.